A conical drinking cup is made from a circular piece of paper of radius R inches by cutting out a sector and joining the edges CA and CB. Find the radius, height and volume of the cone of greatest volume that can be made this way
To find the conical drinking cup of greatest volume, we need to use optimization techniques. Let's break down the problem step by step:
Step 1: Setup the variables
Let r be the radius of the cone, h be the height of the cone, and V be the volume of the cone.
Step 2: Set up the equations
The volume of a cone is given by the formula: V = (1/3) * π * r^2 * h.
Step 3: Relate the variables
To relate the variables r and h, we can use similar triangles. Since the circular piece of paper is the base of the cone, its circumference is equal to the circumference of the cone's base. The circumference of the circular piece of paper is given by: C = 2πR.
The length of the arc AB is equal to the circumference of the base of the cone, which is 2πr. Therefore, the angle of the sector is given by: θ = (2πr / 2πR) * 360° = (r/R) * 360°.
The slant height of the cone can be found using the Pythagorean theorem: l = √(r^2 + h^2).
Step 4: Express the volume in terms of a single variable
Using the equations above, we can express the volume V in terms of a single variable r. Let's substitute the value of h from the Pythagorean theorem equation into the volume formula:
V = (1/3) * π * r^2 * √(r^2 + h^2)
V = (1/3) * π * r^2 * √(r^2 + (l^2 - r^2)) (substituting l^2 - r^2 for h^2)
V = (1/3) * π * r^2 * √(l^2)
V = (1/3) * π * r^2 * l
Step 5: Find the derivative and critical points
To find the critical points, we need to take the derivative of V with respect to r and set it equal to zero. Differentiating V, we get:
dV/dr = (1/3) * π * (2r * l + r^2 * dl/dr)
Setting dV/dr equal to zero:
0 = (1/3) * π * (2r * l + r^2 * dl/dr)
Step 6: Solve for the critical points
To find the critical points, we solve the equation:
2r * l + r^2 * dl/dr = 0
Step 7: Solve for l
To solve for l, we use the equation:
l = √(r^2 + h^2)
Step 8: Use constraints to solve for r and h
To use the constraints, we consider the circular sector length AB:
2πr = (r/R) * 2πR * 360°
2πr = (r/R) * 2πR * π/180 [Using the conversion factor for degrees to radians]
Simplifying, we get:
2r = (r/R) * R * π/180
2 = (r/R) * π/180
Step 9: Solve for r
Solving for r from the constraint equation:
2 = (r/R) * π/180
r/R = 360°/π
r = R * 360°/π
Step 10: Substitute r into the equation found in Step 6 to solve for l
Now that we have r in terms of R, we can substitute it into the equation from Step 6 to solve for l.
2r * l + r^2 * dl/dr = 0
2(R * 360°/π) * l + (R * 360°/π)^2 * dl/dr = 0
Substituting the value of l from Step 7:
2(R * 360°/π) * √(R^2 * (360°/π)^2 + h^2) + (R * 360°/π)^2 * dl/dr = 0
Simplifying further, we get an equation in terms of h:
2(R * 360°/π) * √(R^2 * (360°/π)^2 + h^2) + (R * 360°/π)^2 * dl/dr = 0
Step 11: Solve for h
Now we solve the equation above for h. This may involve using a numerical or graphical method since the equation is transcendental.
Step 12: Plug the values of r and h into the volume formula to find the maximum volume
Finally, plug the values of r and h obtained from Steps 9 and 11, respectively, into the volume formula:
V = (1/3) * π * r^2 * h
This will give you the maximum volume of the conical drinking cup.
To find the radius, height, and volume of the cone of greatest volume that can be made from a circular piece of paper, we need to use calculus and optimization methods.
Let's determine the radius, height, and volume systematically:
1. Let the sector's central angle be θ degrees. The remaining angle of the sector is 360 - θ degrees.
2. After cutting out the sector, the piece of paper will be shaped like a sector of a larger circle.
3. The larger circle's radius is R inches (given in the problem).
4. After cutting out the sector, the cone's radius will be r inches, and the cone's height will be h inches.
5. Since we are looking for the cone of greatest volume, we need to maximize the volume of the cone, V.
Now, let's derive the formulas for the radius, height, and volume of the cone:
1. Radius (r):
- The circumference of the larger circle is 2πR inches.
- The circumference of the cone's base is 2πr inches.
- Since the sector is cut out and joined to form the cone, the arc length CA + arc length CB is equal to the circumference of the larger circle.
- This can be written as (θ/360) * 2πR = 2πr.
- Simplifying the equation, we get r = (θ/360) * R.
2. Height (h):
- The length of the arc AB, which forms the slant height of the cone, is 2πr inches.
- The slant height of the cone can be found using the Pythagorean theorem.
- It can be written as (h^2) + (r^2) = (2πr)^2.
- Simplifying the equation, we get h = √[(2πr)^2 - r^2].
3. Volume (V):
- The volume of a cone is given by the formula V = (1/3) * π * r^2 * h.
- Substituting the values of r and h derived above, we get V as a function of θ, R, and π.
- V = (1/3) * π * [(θ/360) * R]^2 * √[(2π * (θ/360) * R)^2 - [(θ/360) * R]^2].
Now, using calculus methods, we must find the value of θ that maximizes V. We differentiate V with respect to θ, set it equal to 0, and solve for θ. Once we find the value of θ, we can substitute it back into the formulas for r, h, and V to obtain the desired values. Since this process involves advanced calculus, it is recommended to use computational tools or software to find the optimal value.
By following these steps, we can determine the radius, height, and volume of the cone of greatest volume that can be made from the given circular piece of paper.
v = (1/3) base area * height for anything pointy with straight sides
slant height = R
cut away sector angle a
sector angle remaining = 2pi-a
base radius remaining =r= 2piR(2pi-a)/2pi
= R(2 pi-a)
height h = sqrt(R^2-r^2)
v = (1/3)(pi r^2)h
plug and chug